Week 3 Assignment Gough

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Grantham University *

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Electrical Engineering

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Jan 9, 2024

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Kaila Gough G0021 Grantham University 10-01-2023

6. The following resistors (one each) are connected in a series circuit: 1.0 Ω, 2.2 Ω, 5.6 Ω, 12 Ω, and 22Ω. Determine the total resistance. Solve this problem using Rt = R1 + R2 + R3 formula Total resistance equals 42.8Ω 8. Calculate RT for each circuit of Figure 5-66. To solve for total resistance simply add each of the resistor amounts within the circuit. 1.0 + 5.6 + 2.2 = 8.8Ω 10. Six 56Ω resistors, eight 100Ω resistors, and two 22Ω resistors are all connected in series. What is the total resistance? To find the total amount of resistance you must multiply the resistance amounts by the number of resistors there are and then add each of the totals. Formula looks like this: Rt = 6 x 56 + 8 x 100 + 2 x 22 Rt = 336 + 800 + 44 Rt = 1180 ohms 12. You have the following resistor values available to you in the lab in unlimited quantities: 10Ω, 100Ω, 470Ω, 560Ω, 680Ω, 1.0kΩ, 2.2kΩ, and 5.6kΩ. All the other standard values are out of stock. A project that you are working on requires an 18kΩ resistance. What combinations of the available values would you use in series to achieve this total resistance? 10+10+10+10+10+10+10+10+10+10+100+100+100+100+100+100+100+100+100+100+470+470+470+47 0+470+470+470+470+560+560+560+560+560+560+560+560+680+680+680+680 1.0k x 9 2.2k x 4 5.6k x 1 16. The current from the source in Figure 5-69 is 5mA. How much current does each milliammeter in the circuit indicate? To find the current through each resistor we can use formula: I = V/R We know that the voltage source vS is 10v., and the resistors R1 and R2 have values of 100 ohms and 200 ohms respectively. Therefore, the current through jR1 is: I1 = Vs/R1 = 10/100 = 0.1A = 100mA Current through R2 is: I2 = Vs/R2 = 10/200 = 0.05A = 50mA

Since the ammeters are connected in series with the resistors, they will measure the same current as the resistors. Therefore, the ammeter mA connected to R1 will indicate 100mA, and the ammeter connected to R2 will indicate f50mA. The total current form the source is equal to the sum of the currents through each resistor. Therefore, the total current from the source is: I total = I1+ I2 = 100 +50 = 150 mA This means that the current form the source is not accurate as it shows; only 5 mA instead of 150 mA. There may be errors in the circuit or the meter. 20. Determine the voltage drop across each resistor in Figure 5—70. We can solve this using Kirchhoff’s voltage law which is: V = IR So 5.5 = I (2.2) will give the total current since it is a series circuit A. I = 2.5A 2.5(2.2) = 5.5 therefore, R1 = 5.5v 2.5(5.6) = 14v Therefore , R2 = 14. 2.5(1.0) = 2.5 therefore, R3 is 2.5v. B. First, we need to calculate the total resistance which we can do by simply adding each resistance value to come to a total of 382.9787 kΩ. Next, we need to calculate the current flowing through the resistor. Using ohms law, I = V/R The voltage across each of the resistors is 12V. When substituting this value in with the resistance values in ohms law formula we have R1 = 0.012(1000000) = 12v R2 = 0.021(560000) = 11.76V R3 = 0.005(2200000) = 11v Therefore, the voltage drop across each resistor is approximately 12v, 11.76V, and 11v for resistor R1, R2, and R3. 22. Four equal-value resistors are in series with a 5V battery, and 2.23 mA are measured. What is the value of each resistor? To find the value of each resistor we can use ohm’s law for the total resistance of resistors in series. Given by: Rt = R1 +R2 + R3 + R4 since the resistors all have equal values, we can simply this expression to Rt = 4R use ohms law to calculate the current in the circuit: I = V/R the expression is then: 2.23mA = 5v/4R solve for R: R = 5v/(2.23mA*4) = 560Ω Each resistor has a value of 560 Ω. 24. Determine VR1, R2, and R3 in Figure 5-72. First the total R = 382.9787 kΩ current is 12v so our formula is R1 = 12/1000000 = 0.012mA R2 = 12/560000 = 0.021mA R3 = 12/2200000 = 0.005mA therefore the voltage drop across each resistor is given by solving: R1 = 0.012 * 1000000 = 12v R2 = 0.021 * 560000 = 11.76v R3 = 0.0005 *2200000 = 11v. So our voltage drop for each resistor is R1 = 12v, R2 = 11.76v, and R3 = 11v. 30. The term series opposing means that sources are in series with opposite polarities. If a 12V and a 3V battery are series opposing, what is the total voltage. If they are connected in series opposing the voltage of the two batteries will subtract from each other. So 12v – 3v = 9v so the voltage is 9v. Section 5-6 Kirchoff’s Voltage Law 32. The following voltage drop are measured across three resistors in series: 5.5V, 8.2V, and 12.3V. What is the value of the source voltage to which these resistors are connected? To find the source voltage you can simply just add them together so 5.5 + 8.2 + 12.3 = 26v so the source voltage is 26v.

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36. Find R1, R2, and R3 in figure 5-80. R1 = 1.02kΩ, R2= 1.5kΩ, and R3 = 1kΩ. 40. Determine the voltage between points A and B in each voltage divider of Figure 5-82. To determine the voltage divider, we first must figure out the voltage drop across each resistor. The V divider formula is Vout = Vin (R2/(R1+R2) So for the first voltage divider we have Vout = 12v (56/(100+56) Vout = 4.3v therefore the first voltage divider between A and B is 4.3v. For the second we have: Vout = 12 = 12v(100/ (100+1000) Vout = 1.1v therefore our 2 nd voltage divider between A and B is 1.1v. 48. Five series resistors each handle 50mW. What is the total power? When reisstors are connected in series the total power is equaled to the sum of each resistor. So the total power is equal to 5 times 50mW, which is 250mW or 0.25W. 50. If the total resistance of a circuit is halved, what happens to the power? When the total resistance is halved , the power in a circuit quadruples. This is because the power in a circuit is proportional to the square of the current and the resistance. Since it is proportional to the square of the current, it increases by a factor of four. 54. A certain series circuit consists of a 1/8 W resistor, a ¼ W resistor, and a ½ W resistor. The total resistance is 2,400 Ω. If each of the resistors is operating in the circuit at its maximum power dissipation, determine the following: We can use the following formula: P = IV = I^2R I = P/R = (P1+P2+P3)/R)=(0.125+0.25+0.5)/ (2400)=0.025A Vt = IR = (0.025)(2400) = 60V Now we can use ohms law to solve for the value of each resistor: R1 = P1/I^2 = (0.125)/(0.025)^2 = 200ohms R2 = P2/I^2 = (0.25)/(0.025)^2 = 400ohms R3 = P3/I^2 = (0.5)/(0.025)^2 = 800ohms. a. I = 0.025A B. Vt = 60V c. The value of each resistor is 200ohms, 400ohms, and 800ohms.